hi , i recently found while playing around with my new calculator that any number repeated six times is divisible by 7 .... for eg, 4 repeated 6 times gives 444444/7 =63492 an say 761 repeated 6 times 761761761761761761/ 7 =108823108823108823 the last one was done using a computer ... can anyone plz help me out why this happens??? regrds Mahesh
227153715052227153715052227153715052227153715052227153715052227153715052 is 227153715052 repeated 6 times and is not a multiple of 7 as its prime factors are: 2 ^ 2 x 3 x 11 x 19 x 37 x 67 x 73 x 137 x 3169 x 52579 x 98641 x 333667 x 2082527 x 99990001 x 999999000001 x 3199044596370769 Edit: Sorry got a little carried away on my computer
hi , repeating an integer, x (say), six times is not the same as multiplying it six times .... i m sorry but i think u havent got my question correctly... it is like this suppose x is an n digit number repeating it 6 times gives a number y= 1*x + 10^(n)*x + 10^(2n)*x + 10^(3n)*x+ ....+10^(6n)*x which is not equal to x^6 and i think i have got the solution for this problem.... if u want me to post it plz tell me , or if u want to find it urself then its good for u regards Mahesh
well, it's easy to prove for any one digit number n: 10^5*n+10^4*n+10^3*n+10^2*n+10*n+n That is a single digit n repeated 6 times, divide by 7 and you get: 14285 5/7 * n + 1428 4/7 * n + 142 6/7 * n + 14 2/7 * n + 1 3/7 * n + 1/7 * n = 15870*n + (21/7)*n = 15873 * n So, a single digit repeated 6 times will always divide by 7 to an integer. In other words, if you multiply 15873 * 7 * n , where n is a single digit, then you will get a number that is that digit repeated 6 times. If you feel like it, you can attempt similar proofs for more than one digit to see if it holds for more than one digit repeated, and if not than maybe you can find certain conditions for which it would hold with bigger numbers. Or maybe someone has a better way to prove this than me ...
It does not appear to be true for 123423123423123423123423123423123423 either (that is 123423 repeated 6 times).
Repeating a string, s, of r digits six times is the sam as the sum of a geometric progression. Let k be 10^r, it is s +ks +k^2s+..+k^5s The sum of this is (s/9)*(10^6-1) since 9=2 mod 7, and thus invertible, and 10 is coprime with 7, this reduces to zero mod 7 quite trivially by fermat's little theorem.
You do realize this post doesn't make much sense, do you? First, the person you responded to didn't multiple a number 6 times, he/she did what you did, and just repeated one. And your formula is wrong. It should only go up to 10^(5n).
Matt, could you explain this snippet a bit more for me? I've done some modular arithmatic to determine factorability, but this loses me. Thanks.
227153715052227153715052227153715052227153715052227153715052227153715052 is 227153715052 repeated 6 times and not (227153715052)^{6} and it is not a multiple of 7.
Either you or I have made a mistake, then, Zurtex. Now, if there truly were counter examples, then there would be a smaller on than the one you've just written. Incidentally, what is the remainder of that number mod 7? As for dividing by 9. suppose x/9 = y in the integers, then x=9y, and x=9y mod any integer, n say. If 9 and n are relativeyl prime then there is an integer k such that 9k=1 mod n thus xk=y mod n. 9 is a unit modulo 9, 9 is invertible. If 9 and 7 weren't coprime then I couldn't do this.
okay, I followed everything you said individually, but I don't understand "invertible". I'm missing some piece of the puzzle here. Sorry for being dense. Maybe you could write it out in few steps starting from s/9 * (10^6 -1) I would start like this myself (which seems to be more complicated and possibly wrong): 10^6=10^3*10^3 and 10^3 mod 7 = -1 so the whole thing mod 7 = s/9(-1*-1 -1) = 0
Your geometric series should be s/(k-1)*(k^6-1) k-1 won't be invertible mod 7 if r=0 mod 6 (Fermat's little theorem). Note Zurtex example has r=12 and Muzza has r=6. r=6 is the least number of digits you'll find for a counterexample.
I think mahesh really meant the following as he himself corrected: But in any case, what range of number string lengths can be repeated six times and be divisible by 7?
I believe Shmoe's post was pretty clear.Basically any natural number except natural multiples of 6. Daniel.
Is this because 10^6 mod 7 = 1 so that k-1=0? Can someone give me a really simple definition of "invertible" here? Thanks.
I've not been able to keep up the thread but I think you're in the wrong matt: 227153715052227153715052227153715052227153715052227153715052227153715052 mod 7 = 2 There will be smaller counter examples but that's the 1st one I found (mind you I only searched for about 10 secs.
hai , u ppl were right.... when i found one of the solutions for the problem ... i thought it was quite an easy one ... and u rightly pointed out the mistake i made in the formula ... i apologize for anything from my side... This cannot be called a proof.... i noted that (1/7) = 0.142857142857142857..... with '142857' reccuring so y= 1*x + 10^(n)*x + 10^(2n)*x + 10^(3n)*x+ 10^(4n)*x +10^(5n)*x =x(1+10^n+10^2n+10^3n+10^4n+10^5n) Now i multiplied both sides with 0.142857142857..... and found that RHS is always an integer irrespective of the values of x and hence n. thank u guys for helping me cheers Mahesh